Oxidation and Reduction



Oxidation and Reduction

Oxidation means losing electrons and reduction means gaining electrons.

In the reaction between iron and copper(II) sulfate

Fe(s)  +  Cu2+(aq)   [pic]   Fe2+(aq)  +  Cu(s)

Iron metal (Fe(s)) loses 2 electrons to form iron ions (Fe2+(aq))

Fe(s)  -  2e-(aq)   [pic]   Fe2+(aq)

This is called oxidation. Iron is said to be oxidized.

Copper ions (Cu2+(aq)) gain 2 electrons to form copper metal (Cu(s))

Cu2+(aq)  +  2e-(aq)   [pic]   Cu(s)

This is called reduction. Copper is said to be reduced.

To help you to remember that oxidation means losing electrons

and reduction means gaining electrons,

remember  O I L R I G

(Oxidation Is Loss, Reduction Is Gain - of electrons).

Oxidation and reduction always occur together.

They are called redox reactions

(pronounced "reed - ox" for reduction / oxidation).

Oxidation Numbers & Redox Reactions

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The oxidation number of an element indicates the number of electrons lost, gained, or shared as a result of chemical bonding. The change in the oxidation state of a species lets you know if it has undergone oxidation or reduction.

Oxidation can be defined as "an increase in oxidation number".

In other words, if a species starts out at one oxidation state and ends up at a higher oxidation state it has undergone oxidation.

Conversely,

Reduction can be defined as "a decrease in oxidation number".

Any species whose oxidation number is lowered during the course of a reaction has undergone reduction.

Example:

• Na  +  Cl2 ----->  2NaCl

• The Na starts out with an oxidation number of zero (0) and ends up having an oxidation number of 1+. It has been oxidized from a sodium atom to a positive sodium ion.

• The Cl2 also starts out with an oxidation number of zero (0), but it ends up with an oxidation number of 1-. It, therefore, has been reduced from chlorine atoms to negative chloride ions.

The substance bringing about the oxidation of the sodium atoms is the chlorine, thus the chlorine is called an oxidizing agent. In other words, the oxidizing agent is being reduced (undergoing reduction).

The substance bringing about the reduction of the chlorine is the sodium, thus the sodium is called a reducing agent. Or in other words, the reducing agent is being oxidized (undergoing oxidation).

Oxidation is ALWAYS accompanied by reduction. Reactions in which oxidation and reduction are occurring are usually called Redox reactions.

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Rules for Assigning Oxidation Numbers

There are several rules for assigning the oxidation number to an element. Learning these rules will simplify the task of determining the oxidation state of an element, and thus, whether it has undergone oxidation or reduction.

1. The oxidation number of an atom in the elemental state is zero.

Example: Cl2 and Al both are 0

2. The oxidation number of a monatomic ion is equal to its charge.

Example: In the compound NaCl, the sodium has an oxidation number of 1+ and the chlorine is 1-.

3. The algebraic sum of the oxidation numbers in the formula of a compound is zero.

Example: the oxidation numbers in the NaCl above add up to 0

4. The oxidation number of hydrogen in a compound is 1+, except when hydrogen forms compounds called hydrides with active metals, and then it is 1-.

Examples: H is 1+ in H2O, but 1- in NaH (sodium hydride).

5. The oxidation number of oxygen in a compound is 2-, except in peroxides when it is 1-, and when combined with fluorine. Then it is 2+.

Example: In H2O the oxygen is 2-, in H2O2 it is 1-.

6. The algebraic sum of the oxidation numbers in the formula for a polyatomic ion is equal to the charge on that ion.

Example: in the sulfate ion, SO42-, the oxidation numbers of the sulfur and the oxygen’s add up to 2-. The oxygen’s are 2- each, and the sulfur is 6+.

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Application Problems:

1. What is the oxidation number of chromium in

Na2CrO4

Cr2O72-

2. Given the unbalanced equation below:

Cr2O3(s)  +  Al(s) ---->  Cr(s)  +  Al2O3(s)

a. identify the oxidation state of each element

b. identify the oxidizing agent

c. identify the reducing agent

Balancing Redox Reactions

Redox reactions often can be most easily balanced by

a. first dividing them into half-reactions; that is, the oxidation reaction and the reduction reaction.

Cr3+(aq)  +  Cl1-(aq) ---->  Cr(s)  +  Cl2(g)

Oxidation:  Cl1-(aq) ---->  Cl2(g)  +  2e-

Reduction:  Cr3+(aq)  +  3e- ---->  Cr(s)

b. balance each half-reaction with respect to atoms first, then with respect to electrons.

Oxidation:  3( 2Cl1-(aq) ---->  Cl2(g) +   2e- )

Reduction:  2( Cr3+(aq)  +  3e- ---->  Cr(s) )

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Oxidation:  6Cl1-(aq) ---->  3Cl2(g)  +  6e-

Reduction:  2Cr3+(aq)  +   6e- ---->  2Cr(s)

c. add the two half-reactions together canceling the electrons which are now equal on each side of the arrow.

Final Equation:  2Cr3+(aq)  +  6Cl1-(aq) ----->  2Cr(s)  +  3Cl2(g)

Granted, this example is a simple one, however, the same basic procedure is carried out for most redox reactions, with the addition of a couple of other steps, depending on the reaction conditions.

Example:  copper metal added to concentrated nitric acid

Cu(s)  +  HNO3(aq) ----->  Cu(NO3)2(aq)  +  NO2(g)

a. first divide the equation into half-reactions (notice that the copper is going from a 0 oxidation state to 2+ which is oxidation, and some of the nitrogen is being reduced from 5+ in the nitrate ion to 4+ in the nitrogen dioxide).

Oxidation:  Cu(s) ----->  Cu2+(aq)  +  2e-

Reduction:  NO31-(aq)  +  e- ----->  NO2(g)

b. balance each half-reaction with respect to atoms first, then with respect to electrons.

Oxidation:  Cu(s) ----->  Cu2+(aq)  +  2e-

Reduction:  2NO31-(aq)  +  2 e- ----->  2NO2(g))

Notice that this time there are oxygen atoms in the reduction step that are not balanced. When this happens, add as many water molecules on the right as are needed to balance the total oxygens on the left. Then add hydrogen ions on the left side of the arrow to balance the number of hydrogen atoms that were introduced with the water molecules.

Oxidation:  Cu(s) ----->   Cu2+(aq)  +  2e-

Reduction:  2NO31-(aq)  +  2 e-  +  4H1+(aq) ----->  2NO2(g))  +  2H2O(l)

c. add the two half-reactions together canceling the electrons which are now equal on each side of the arrow.

Cu(s)  +  2NO31-(aq)  +  4H1+(aq) ----->  Cu2+(aq)  +  2NO2(g))  +   2H2O(l)

Note that this is a Net Ionic Equation for the reaction. The other two nitrate ions that would be with the four hydrogen ions in the nitric acid would remain unchanged and provide the nitrate ions that would form copper(II) nitrate.

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